Question

Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.

Answer 1

Let x be the smaller of the two consecutive odd positive integers. Then, the other integer is 

x+ 2. 
Since both the integers are smaller than 10,
x+ 2 < 10 
⇒ x < 10-2 
⇒ x < 8 ..… (i)
Also, the sum of the two integers is more than 11.
∴ x+ (x + 2) > 11 
⇒ 2x+ 2 > 11
⇒ 2x > 11-2 
⇒ 2x > 9
\(⇒ x > \frac{9}{2}\)
⇒ x > 4.5 ..... (ii)
From (i) and (ii), we obtain
Since x is an odd number, x can take the values, 5 and 7.
Thus, the required possible pairs are (5, 7) and (7, 9).